y=tan^-1((1+x)/(1-x)) হলে dy/dx=?

Explanation:

Another Explanation (5): দেওয়া আছে, \(y = \tan^{-1}\left(\frac{1+x}{1-x}\right)\). 🤔 ধরি, \(x = \tan\theta\). সুতরাং, \(\theta = \tan^{-1}x\). 🤓 তাহলে, \(y = \tan^{-1}\left(\frac{1+\tan\theta}{1-\tan\theta}\right)\). আমরা জানি, \(\tan\left(\frac{\pi}{4} + \theta\right) = \frac{1+\tan\theta}{1-\tan\theta}\). 🎉 সুতরাং, \(y = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + \theta\right)\right) = \frac{\pi}{4} + \theta\). 🥳 যেহেতু \(\theta = \tan^{-1}x\), তাই \(y = \frac{\pi}{4} + \tan^{-1}x\). 🤩 এখন, \(x\) এর সাপেক্ষে অন্তরকলন করে পাই, \(\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4} + \tan^{-1}x\right)\). 😇 \(\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4}\right) + \frac{d}{dx}\left(\tan^{-1}x\right)\). 😉 আমরা জানি, \(\frac{d}{dx}\left(\tan^{-1}x\right) = \frac{1}{1+x^2}\) এবং \(\frac{d}{dx}(c) = 0\), যেখানে \(c\) একটি ধ্রুবক। 😎 সুতরাং, \(\frac{dy}{dx} = 0 + \frac{1}{1+x^2} = \frac{1}{1+x^2}\). ✨ অতএব, \(\frac{dy}{dx} = \frac{1}{1+x^2}\). 💖